basic probability

[ignacio_buendia]

Is it correct to state the following?

If the difficulty of the protocol requires a digest with 5 leading zeros, than I have the chance of ((1/16^5)+(1/16^6)+...(1/16^32)) to find a valid proof of work with 1 attempt - regardless of any variables?

[j2002ba2]

No. The zeroes are independent. The probability of having 5 (hexadecimal) leading zeroes is 16^-5.

[BrewMaster]

since the difficulty is defined as an integer comparison not a "leading zero count" thing, if you want to compute the probability of finding a block at first attempt then you must find the probability of finding a number that is in range from 0 to target from a bigger range of numbers that go from 0 to 2^256.
and if i'm not mistaken that probability is calculated by dividing target by 2^256.